4.10e Second order non-homogeneous: complementary + particular integral

2. The velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) of a particle \(P\) at time \(t\) seconds satisfies the vector differential equation $$\frac { \mathrm { d } \mathbf { v } } { \mathrm {~d} t } + 4 \mathbf { v } = \mathbf { 0 }$$ The position vector of \(P\) at time \(t\) seconds is \(\mathbf { r }\) metres.
Given that at \(t = 0 , \mathbf { r } = ( \mathbf { i } - \mathbf { j } )\) and \(\mathbf { v } = ( - 8 \mathbf { i } + 4 \mathbf { j } )\), find \(\mathbf { r }\) at time \(t\) seconds.

2. At time \(t\) seconds, the position vector of a particle \(P\) is \(\mathbf { r }\) metres, where \(\mathbf { r }\) satisfies the vector differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \mathbf { r } = \mathrm { e } ^ { 2 t } \mathbf { j }$$ When \(t = 0 , P\) has position vector \(( \mathbf { i } + \mathbf { j } ) \mathrm { m }\) and velocity \(2 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find an expression for \(\mathbf { r }\) in terms of \(t\).

1. A particle \(P\) is moving in a plane. At time \(t\) seconds the position vector of \(P\) is \(\mathbf { r }\) metres and the velocity of \(P\) is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\). When \(t = \frac { \pi } { 2 } , P\) is instantaneously at rest at the point with position vector \(( \mathbf { i } - \mathbf { j } ) \mathrm { m }\).

Given that \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \mathbf { r } = ( 3 \sin t ) \mathbf { i }$$ find \(\mathbf { v }\) in terms of \(t\).
(13)

15 In an oscillating system, a particle of mass \(m \mathrm {~kg}\) moves in a horizontal line. Its displacement from its equilibrium position O at time \(t\) seconds is \(x\) metres, its velocity is \(v \mathrm {~ms} ^ { - 1 }\), and it is acted on by a force \(2 m x\) newtons acting towards O as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{b57a2590-84e8-4998-9633-902db861f23a-6_212_914_408_242} Initially, the particle is projected away from O with speed \(1 \mathrm {~ms} ^ { - 1 }\) from a point 2 m from O in the positive direction.

1. 1. Show that the motion is modelled by the differential equation \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 x = 0\).
   2. State the type of motion.
   3. Write down the period of the motion.
   4. Find \(x\) in terms of \(t\).
   5. Find the amplitude of the motion.
2. The motion is now damped by a force \(2 m v\) newtons.
   1. Show that \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 \frac { d x } { d t } + 2 x = 0\).
   2. State, giving a reason, whether the system is under-damped, critically damped or over-damped.
   3. Determine the general solution of this differential equation.
3. Finally, a variable force \(2 m \cos 2 t\) newtons is added, so that the motion is now modelled by the differential equation \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 \frac { d x } { d t } + 2 x = 2 \cos 2 t\).
   1. Find \(x\) in terms of \(t\). In the long term, the particle is seen to perform simple harmonic motion with a period of just over 3 seconds.
   2. Verify that this behaviour is consistent with the answer to part (c)(i).

14

1. Find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } - 2 y = 12 e ^ { - x }\). You are given that \(y\) tends to zero as \(x\) tends to infinity, and that \(\frac { \mathrm { dy } } { \mathrm { dx } } = 0\) when \(x = 0\).
2. Find the exact value of \(x\) for which \(y = 0\).

13 Find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } + 2 \frac { d y } { d x } - 3 y = 2 e ^ { x }\).

6. The diagram on the left shows a train of mass 50 tonnes approaching a buffer at the end of a straight horizontal railway track. The buffer is designed to prevent the train from running off the end of the track. The buffer may be modelled as a light horizontal spring \(A B\), as shown in the diagram on the right, which is fixed at the end \(A\). The train strikes the buffer so that \(P\) makes contact with \(B\) at \(t = 0\) seconds. While \(P\) is in contact with \(B\), an additional resistive force of \(250000 v \mathrm {~N}\) will oppose the motion of the train, where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of the train at time \(t\) seconds. The spring has natural length 1 m and modulus of elasticity 312500 N . At time \(t\) seconds, the compression of the spring is \(x\) metres. \includegraphics[max width=\textwidth, alt={}, center]{d7f600c5-af4a-4708-bfd9-92b37a95c634-7_358_1506_824_283}

1. Show that, while \(P\) is in contact with \(B\), \(x\) satisfies the differential equation $$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 20 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 25 x = 0$$
2. Given that, when \(P\) first makes contact with \(B\), the speed of the train is \(U \mathrm {~ms} ^ { - 1 }\), find an expression for \(x\) in terms of \(U\) and \(t\).
3. When the train comes to rest, the compression of the buffer is 0.3 m . Determine the speed of the train when it strikes the buffer.
4. State which type of damping is described by the motion of \(P\). Give a reason for your answer.

1. A scientist is studying the effect of introducing a population of white-clawed crayfish into a population of signal crayfish.
   At time \(t\) years, the number of white-clawed crayfish, \(w\), and the number of signal crayfish, \(s\), are modelled by the differential equations

$$\begin{aligned} & \frac { \mathrm { d } w } { \mathrm {~d} t } = \frac { 5 } { 2 } ( w - s ) \\ & \frac { \mathrm { d } s } { \mathrm {~d} t } = \frac { 2 } { 5 } w - 90 \mathrm { e } ^ { - t } \end{aligned}$$

1. Show that $$2 \frac { \mathrm {~d} ^ { 2 } w } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} w } { \mathrm {~d} t } + 2 w = 450 \mathrm { e } ^ { - t }$$
2. Find a general solution for the number of white-clawed crayfish at time \(t\) years.
3. Find a general solution for the number of signal crayfish at time \(t\) years. The model predicts that, at time \(T\) years, the population of white-clawed crayfish will have died out. Given that \(w = 65\) and \(s = 85\) when \(t = 0\)
4. find the value of \(T\), giving your answer to 3 decimal places.
5. Suggest a limitation of the model.

1. Two compounds, \(X\) and \(Y\), are involved in a chemical reaction. The amounts in grams of these compounds, \(t\) minutes after the reaction starts, are \(x\) and \(y\) respectively and are modelled by the differential equations

$$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = - 5 x + 10 y - 30 \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = - 2 x + 3 y - 4 \end{aligned}$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 50$$
2. Find, according to the model, a general solution for the amount in grams of compound \(X\) present at time \(t\) minutes.
3. Find, according to the model, a general solution for the amount in grams of compound \(Y\) present at time \(t\) minutes. Given that \(x = 2\) and \(y = 5\) when \(t = 0\)
4. find
   1. the particular solution for \(x\),
   2. the particular solution for \(y\). A scientist thinks that the chemical reaction will have stopped after 8 minutes.
5. Explain whether this is supported by the model.

10. \begin{figure}[h] \end{figure} The motion of a pendulum, shown in Figure 3, is modelled by the differential equation $$\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 9 \theta = \frac { 1 } { 2 } \cos 3 t$$ where \(\theta\) is the angle, in radians, that the pendulum makes with the downward vertical, \(t\) seconds after it begins to move. \begin{enumerate}[label=(\alph*)] \item

1. Show that a particular solution of the differential equation is $$\theta = \frac { 1 } { 12 } t \sin 3 t$$
2. Hence, find the general solution of the differential equation. Initially, the pendulum
   Given that, 10 seconds after it begins to move, the pendulum makes an angle of \(\alpha\) radians with the downward vertical,
3. determine, according to the model, the value of \(\alpha\) to 3 significant figures. Given that the true value of \(\alpha\) is 0.62
4. evaluate the model. The differential equation $$\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 9 \theta = \frac { 1 } { 2 } \cos 3 t$$ models the motion of the pendulum as moving with forced harmonic motion.
5. Refine the differential equation so that the motion of the pendulum is simple harmonic motion.

1. A company plans to build a new fairground ride. The ride will consist of a capsule that will hold the passengers and the capsule will be attached to a tall tower. The capsule is to be released from rest from a point half way up the tower and then made to oscillate in a vertical line.

The vertical displacement, \(x\) metres, of the top of the capsule below its initial position at time \(t\) seconds is modelled by the differential equation, $$m \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 200 \cos t , \quad t \geqslant 0$$ where \(m\) is the mass of the capsule including its passengers, in thousands of kilograms.
The maximum permissible weight for the capsule, including its passengers, is 30000 N .
Taking the value of \(g\) to be \(10 \mathrm {~ms} ^ { - 2 }\) and assuming the capsule is at its maximum permissible weight,

1. 1. explain why the value of \(m\) is 3
   2. show that a particular solution to the differential equation is $$x = 40 \sin t - 20 \cos t$$
   3. hence find the general solution of the differential equation.
2. Using the model, find, to the nearest metre, the vertical distance of the top of the capsule from its initial position, 9 seconds after it is released.

1. A scientist is investigating the concentration of antibodies in the bloodstream of a patient following a vaccination.
   The concentration of antibodies, \(x\), measured in micrograms ( \(\mu \mathrm { g }\) ) per millilitre ( ml ) of blood, is modelled by the differential equation

$$100 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 60 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 26$$ where \(t\) is the number of weeks since the vaccination was given.

1. Find a general solution of the differential equation. Initially,
   • there are no antibodies in the bloodstream of the patient
   • the concentration of antibodies is estimated to be increasing at \(10 \mu \mathrm {~g} / \mathrm { ml }\) per week
   • Find, according to the model, the maximum concentration of antibodies in the bloodstream of the patient after the vaccination.
   A second dose of the vaccine has to be given to try to ensure that it is fully effective. It is only safe to give the second dose if the concentration of antibodies in the bloodstream of the patient is less than \(5 \mu \mathrm {~g} / \mathrm { ml }\).
2. Determine whether, according to the model, it is safe to give the second dose of the vaccine to the patient exactly 10 weeks after the first dose.

1. A patient is treated by administering an antibiotic intravenously at a constant rate for some time.

Initially there is none of the antibiotic in the patient.
At time \(t\) minutes after treatment began

• the concentration of the antibiotic in the blood of the patient is \(x \mathrm { mg } / \mathrm { ml }\)
• the concentration of the antibiotic in the tissue of the patient is \(y \mathrm { mg } / \mathrm { ml }\)

The concentration of antibiotic in the patient is modelled by the equations $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.025 y - 0.045 x + 2 \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = 0.032 x - 0.025 y \end{aligned}$$

1. Show that $$40000 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2800 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 13 y = 2560$$
2. Determine, according to the model, a general solution for the concentration of the antibiotic in the patient's tissue at time \(t\) minutes after treatment began.
3. Hence determine a particular solution for the concentration of the antibiotic in the tissue at time \(t\) minutes after treatment began. To be effective for the patient the concentration of antibiotic in the tissue must eventually reach a level between \(185 \mathrm { mg } / \mathrm { ml }\) and \(200 \mathrm { mg } / \mathrm { ml }\).
4. Determine whether the rate of administration of the antibiotic is effective for the patient, giving a reason for your answer.

1. The motion of a particle \(P\) along the \(x\)-axis is modelled by the differential equation

$$2 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x = 4 t + 12$$ where \(P\) is \(x\) metres from the origin \(O\) at time \(t\) seconds, \(t \geqslant 0\)

1. Determine the general solution of the differential equation.
2. Hence determine the particular solution for which \(x = 3\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 2\) when \(t = 0\)
3. 1. Show that, according to the model, the minimum distance between \(O\) and \(P\) is \(( 2 + \ln 2 )\) metres.
   2. Justify that this distance is a minimum. For large values of \(t\) the particle is expected to move with constant speed.
4. Comment on the suitability of the model in light of this information.

1. A particle \(P\) moves along a straight line.

At time \(t\) minutes, the displacement, \(x\) metres, of \(P\) from a fixed point \(O\) on the line is modelled by the differential equation $$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x + 16 t ^ { 2 } x = 4 t ^ { 3 } \sin 2 t$$

1. Show that the transformation \(x =\) ty transforms equation (I) into the equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 16 y = 4 \sin 2 t$$
2. Hence find a general solution for the displacement of \(P\) from \(O\) at time \(t\) minutes.

1. The motion of a particle \(P\) along the \(x\)-axis is modelled by the differential equation

$$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t ( t + 1 ) \frac { \mathrm { d } x } { \mathrm {~d} t } + 2 ( t + 1 ) x = 8 t ^ { 3 } \mathrm { e } ^ { t }$$ where \(P\) has displacement \(x\) metres from the origin \(O\) at time \(t\) minutes, \(t > 0\)

1. Show that the transformation \(x = t u\) transforms the differential equation (I) into the differential equation $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} u } { \mathrm {~d} t } = 8 \mathrm { e } ^ { t }$$ Given that \(P\) is at \(O\) when \(t = \ln 3\) and when \(t = \ln 5\)
2. determine the particular solution of the differential equation (I)

1. A vibrating spring, fixed at one end, has an external force acting on it such that the centre of the spring moves in a straight line. At time \(t\) seconds, \(t \geqslant 0\), the displacement of the centre \(C\) of the spring from a fixed point \(O\) is \(x\) micrometres.

The displacement of \(C\) from \(O\) is modelled by the differential equation $$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t \frac { \mathrm {~d} x } { \mathrm {~d} t } + \left( 2 + t ^ { 2 } \right) x = t ^ { 4 }$$

1. Show that the transformation \(x = t v\) transforms equation (I) into the equation $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} t ^ { 2 } } + v = t$$
2. Hence find the general equation for the displacement of \(C\) from \(O\) at time \(t\) seconds.
3. 1. State what happens to the displacement of \(C\) from \(O\) as \(t\) becomes large.
   2. Comment on the model with reference to this long term behaviour.

8. $$\mathbf { A } = \left( \begin{array} { r r r } 5 & - 2 & 5 \\ 0 & 3 & p \\ - 6 & 6 & - 4 \end{array} \right) \quad \text { where } p \text { is a constant }$$ Given that \(\left( \begin{array} { r } 2 \\ 1 \\ - 2 \end{array} \right)\) is an eigenvector for \(\mathbf { A }\)

1. 1. determine the eigenvalue corresponding to this eigenvector
   2. hence show that \(p = 2\)
   3. determine the remaining eigenvalues and corresponding eigenvectors of \(\mathbf { A }\)
2. Write down a matrix \(\mathbf { P }\) and a diagonal matrix \(\mathbf { D }\) such that \(\mathbf { A } = \mathbf { P D P } ^ { - 1 }\)
3. 1. Solve the differential equation \(\dot { u } = k u\), where \(k\) is a constant. With respect to a fixed origin \(O\), the velocity of a particle moving through space is modelled by $$\left( \begin{array} { c } \dot { x } \\ \dot { y } \\ \dot { z } \end{array} \right) = \mathbf { A } \left( \begin{array} { l } x \\ y \\ z \end{array} \right)$$ By considering \(\left( \begin{array} { c } u \\ v \\ w \end{array} \right) = \mathbf { P } ^ { - 1 } \left( \begin{array} { c } x \\ y \\ z \end{array} \right)\) so that \(\left( \begin{array} { c } \dot { u } \\ \dot { v } \\ \dot { w } \end{array} \right) = \mathbf { P } ^ { - 1 } \left( \begin{array} { c } \dot { x } \\ \dot { y } \\ \dot { z } \end{array} \right)\)
   2. determine a general solution for the displacement of the particle.

6. (a) Determine the general solution of the recurrence relation $$u _ { n } = 2 u _ { n - 1 } - u _ { n - 2 } + 2 ^ { n } \quad n \geqslant 2$$ (b) Hence solve this recurrence relation given that \(u _ { 0 } = 2 u _ { 1 }\) and \(u _ { 4 } = 3 u _ { 2 }\)

1

1. Find the roots of the equation \(m ^ { 2 } + 2 m + 2 = 0\) in the form \(a + i b\).
   (2 marks)
2. 1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 4 x$$
   2. Hence express \(y\) in terms of \(x\), given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2\) when \(x = 0\).

5 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = 6 + 5 \sin x$$ (12 marks)

1

1. Find the value of the constant \(k\) for which \(k x ^ { 2 } \mathrm { e } ^ { 5 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 10 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 25 y = 6 \mathrm { e } ^ { 5 x }$$
2. Hence find the general solution of this differential equation.

10

1. Find the particular solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 10 x = 37 \sin 3 t$$ given that \(x = 3\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\) when \(t = 0\).
2. Show that, for large positive values of \(t\) and for any initial conditions, $$x \approx \sqrt { } ( 37 ) \sin ( 3 t - \phi ) ,$$ where the constant \(\phi\) is such that \(\tan \phi = 6\).

1. The vertical height, \(h \mathrm {~m}\), above horizontal ground, of a passenger on a fairground ride, \(t\) seconds after the ride starts, where \(t \leqslant 5\), is modelled by the differential equation

$$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } h } { \mathrm {~d} t ^ { 2 } } - 2 t \frac { \mathrm {~d} h } { \mathrm {~d} t } + 2 h = t ^ { 3 }$$

1. Given that \(t = \mathrm { e } ^ { x }\), show that
   1. \(t \frac { \mathrm {~d} h } { \mathrm {~d} t } = \frac { \mathrm { d } h } { \mathrm {~d} x }\)
   2. \(t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } h } { \mathrm {~d} t ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } h } { \mathrm {~d} x ^ { 2 } } - \frac { \mathrm { d } h } { \mathrm {~d} x }\)
2. Hence show that the transformation \(t = \mathrm { e } ^ { x }\) transforms equation (I) into the equation $$\frac { \mathrm { d } ^ { 2 } h } { \mathrm {~d} x ^ { 2 } } - 3 \frac { \mathrm {~d} h } { \mathrm {~d} x } + 2 h = \mathrm { e } ^ { 3 x }$$
3. Hence show that $$h = A t + B t ^ { 2 } + \frac { 1 } { 2 } t ^ { 3 }$$ where \(A\) and \(B\) are constants. Given that when \(t = 1 , h = 2.5\) and when \(t = 2 , \frac { \mathrm {~d} h } { \mathrm {~d} t } = - 1\)
4. determine the height of the passenger above the ground 5 seconds after the start of the ride.

8 For the differential equation \(t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 4 t \frac { \mathrm {~d} x } { \mathrm {~d} t } + \left( 6 - 4 t ^ { 2 } \right) x = 0\), use the substitution \(x = t ^ { 2 } u\) to find a differential equation involving \(t\) and \(u\) only. Hence solve the above differential equation, given that \(x = \mathrm { e } - 1\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 4 \mathrm { e }\) when \(t = 1\).